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Let $(X,\mu)$ be a probability measure space and $K:X \times X \to \mathbb R$ be a (psd) kernel on $X$. Let $K_0$ be another kernel on $X$ and defined a new kernel $\widetilde K$ on $X$ by $$ \widetilde K(x,y) = \int_X\int_X K_0(x,a)K(a,b)K_0(b,y)d\mu(a)d\mu(b). $$

(with obvious integrability assumptions)

Question. Can the properties of $\widetilde{K}$ (e.g of its induced RKHS, eigenspectrum of induced kernel integral operator, etc.) be deduced from the properties of $K$ ?

I'm particularly interested in the case where $K$ is an inner-product kernel and

  • $(X,\mu) = (\mathbb R^n,(2\pi)^{-n/2}e^{-\|x\|^2/2}dx)$, where $dx$ is Lebesgue measure in $\mathbb R^n$, OR
  • $(X,\mu) = (S_{n-1},\tau_n)$, where $S_{n-1}$ is the unit-sphere in $\mathbb R^n$ and $\tau_n$ is the uniform distribution thereupon.
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    $\begingroup$ Not sure if this helps or is well-known to you anyway. Let $H$ resp. $\tilde{H}$ the RKHS of $K$ resp. $\tilde{K}$. I would expect that $\tilde{H}=\{\int_X K_0(x,a)f(a)d\mu(a) \mid f\in H\}$ to hold under mild assumptions. At least such results hold for differentiation and several other linear operators, see e.g. arxiv.org/abs/1801.09197 or Lemma 2.2 in arxiv.org/abs/2002.00818. $\endgroup$
    – Markus Lange-Hegermann
    Commented Feb 18, 2022 at 10:48
  • $\begingroup$ It does helps very much. Thanks. $\endgroup$
    – dohmatob
    Commented Feb 21, 2022 at 7:22

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